Central Limit Theorem-CLT

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Presentation transcript:

Central Limit Theorem-CLT MM4D1. Using simulation, students will develop the idea of the central limit theorem.

Central Limit Theorem - CLT The central limit theorem states that the sampling distribution of any statistic will be normal or nearly normal, if the sample size is large enough. Generally, a sample size is considered "large enough" if any of the following conditions apply: The population distribution is normal. The sample distribution is roughly symmetric, unimodal, without outliers, and the sample size is 15 or less. The sample distribution is moderately skewed, unimodal, without outliers, and sample size is between 16 and 40. The sample size is greater than 40, without outliers

CLT-continue  

CLT Formula Use to gain information about a sample mean Use to gain information about an individual data value    

To calculate the CLT  

Examples 1. A bottling company uses a filling machine to fill plastic bottles with a popular cola. The bottles are supposed to contain 300 ml. In fact, the contents vary according to a normal distribution with mean µ = 303 ml and standard deviation σ = 3 ml. What is the probability that an individual bottle contains less than 300 ml? What is the probability that an individual bottle contains greater than 300 ml? What is the probability that an individual bottle contains between 300 ml and 350 ml? Now take a random sample of 10 bottles. What are the mean and standard deviation of the sample mean contents x-bar of these 10 bottles? What is the probability that the sample mean contents of the 10 bottles is less than 300 ml?

Solution a) use z=(x-µ)/σ) & Table of negative Z-score z=(300-303)/3 = -1 P(x<300) = 0.1587 or 15.87%, b) P(x>300) = 1 – 0.1587 = 0.8413 or 84.13% c) z=(x-µ)/σ) z=(300-303)/3 = -1 → P(x=300) = 0.1587 z=(310-303)/3 = 2.33 → P(x=310) = 0.9893 P(300<x<310) = 0.9893-0.1587 = 0.8306 or 83.06% d) mean: 303, stdev: 3/sqrt(10) = 0.94868 e) z=(x-µ)/(σ/sqrt(10) & Table of negative Z-score z=(300-303)/0.94868 = -3.16 p=0.0008 or 0.08%